Q. Find the frequency of minimum distance between compression & rarefaction of a wire. If the length of the wire is $1\ m$ & velocity of sound in air is $360\ m/s$
A) $90\ \sec^{- 1}$
B) $180\ \sec^{- 1}$
C) $120\ \sec^{- 1}$
D) $360\ \sec^{- 1}$
Explanation
(b) The minimum distance between compression and refraction of the wire $l = \frac{\lambda}{2}\ \therefore$ Wave length $\lambda = 2l$
Now by $v = n\lambda \Rightarrow n = \frac{360}{2 \times 1} = 180s{ec}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.